# Proving (cover) compactness implies completeness without using sequential compactness

It is well known that for a subset $X$ in a metric space $M$, the following are equivalent:

(1) Every open cover of $X$ has a finite subcover.

(2) Every sequence in $X$ has a subsequential limit in $X$.

(3) $X$ is complete and totally bounded.

In Marsden's Elementary Classical Analysis 2nd edition, one can see that the proof of "3.1.3 Theorem" shows (1) and (2) imply each other and that the proof of "3.1.5 Theorem" shows (2) and (3) imply each other.

Currently I am trying to prove (1) and (3) imply each other as a (self-studying) exercise.

Currently the most difficult part probably is proving (cover) compactness implies completeness. The uncomplete proof I have written is the following:

Assume for contradiction that $X$ is not complete. That is, assume there exist a sequence $\{x_{n}\}_{n=0}^{\infty}$ in $X$ which (i) is a Cauchy sequence and (ii) does not converge in $X$. (ii) means: for every element $p$ in $X$ there exist $\epsilon_{p} > 0$ such that, for all $N\in\Bbb{N}$ there exists $N'>N$ such that $d(p_{N'},p)\geq\epsilon_{p}$. $\{B_{\epsilon_{p}}(p):p\in{X}\}$ is an open cover of $X$. By (1), there is a finite subcover $\{B_{\epsilon_{p_{1}}}(p_{1}),B_{\epsilon_{p_{2}}}(p_{2}),...,B_{\epsilon_{p_{\alpha}}}(p_{\alpha})\}$.

Now my intuitive idea is that: By (i) there can exists a sufficiently small open ball $B_{*}$ such that (iii) $B_{*}$ is contained in one of the finite subcover balls $B_{\epsilon_{p_{t}}}(p_{t})$ and (iv) for all suffitiently large $n$, $B_{*}$ contains all terms $x_{n}$ of the Cauchy sequence $\{x_{n}\}_{n=0}^{\infty}$. But $\{x_{n}\}_{n=0}^{\infty}$ does not converges to $p_{t}$ so there are infinitely many arbitrarily large $m$'s such that $x_{m}\notin{B_{\epsilon_{p_{t}}}(p_{t})}$ and hence $x_{m}\notin{B_{*}}$, which is a contradiction. Hence $X$ must be complete.

But I have tried very hard but still cannot turn my intuitive idea into a complete rigorous proof. Could someone please give me some instructions?

• Use that a Cauchy sequence converges to all of its accumulation points, hence you can suppose that your Cauchy sequence has no accumulation point. And that means you get balls that contain $x_n$ for only finitely many $n$. – Daniel Fischer Sep 29 '17 at 13:47

Apply compactness to the balls $B_{\varepsilon_p/2}$ to get finitely many $p_1,\ldots,p_\alpha$ whose $\varepsilon_{p_i}/2$ balls cover the space. Let $\varepsilon$ be the minimum of $\varepsilon_{p_1}/2,\ldots,\varepsilon_{p_\alpha}/2$. Since the sequence is Cauchy, for sufficiently large $N$, all but finitely many terms of the sequence are within $\varepsilon$ of $x_N$. Since $x_N$ must be in some ball $B_{\varepsilon_{p_i}/2}$, the triangle inequality gives that eventually all elements of the sequence are in $B_{\varepsilon_{p_i}}$. Contradiction!